Return the maximum sum or -1 if no such pair exists.

Example 1:

Input: nums = [51,71,17,24,42]
Output: 88
Explanation: 
For i = 1 and j = 2, nums[i] and nums[j] have equal maximum digits with a pair sum of 71 + 17 = 88. 
For i = 3 and j = 4, nums[i] and nums[j] have equal maximum digits with a pair sum of 24 + 42 = 66.
It can be shown that there are no other pairs with equal maximum digits, so the answer is 88.

Example 2:

Input: nums = [1,2,3,4]
Output: -1
Explanation: No pair exists in nums with equal maximum digits.

Constraints:

• 2 <= nums.length <= 100
• 1 <= nums[i] <= 104

Solution: Brute Force

Enumerate all pairs of nums and check their sum and max digit.

Time complexity: O(n²)
Space complexity: O(1)

// Author: Huahua
class Solution {
public:
  int maxSum(vector<int>& nums) {
    auto maxDigit = [](int x) {
      int ans = 0;
      while (x) {
        ans = max(ans, x % 10);
        x /= 10;
      }
      return ans;
    };
    int ans = -1;
    const int n = nums.size();
    for (int i = 0; i < n; ++i)
      for (int j = i + 1; j < n; ++j)
        if (nums[i] + nums[j] > ans 
            && maxDigit(nums[i]) == maxDigit(nums[j]))
          ans = nums[i] + nums[j];
    return ans;
  }
};

The post 花花酱 LeetCode 2815. Max Pair Sum in an Array appeared first on Huahua's Tech Road.

1. A decimal number or an integer.
2. (Optional) An 'e' or 'E', followed by an integer.

A decimal number can be split up into these components (in order):

1. (Optional) A sign character (either '+' or '-').
2. One of the following formats:
   1. One or more digits, followed by a dot '.'.
   2. One or more digits, followed by a dot '.', followed by one or more digits.
   3. A dot '.', followed by one or more digits.

An integer can be split up into these components (in order):

1. (Optional) A sign character (either '+' or '-').
2. One or more digits.

For example, all the following are valid numbers: ["2", "0089", "-0.1", "+3.14", "4.", "-.9", "2e10", "-90E3", "3e+7", "+6e-1", "53.5e93", "-123.456e789"], while the following are not valid numbers: ["abc", "1a", "1e", "e3", "99e2.5", "--6", "-+3", "95a54e53"].

Given a string s, return true if s is a valid number.

Example 1:

Input: s = "0"
Output: true

Example 2:

Input: s = "e"
Output: false

Example 3:

Input: s = "."
Output: false

Example 4:

Input: s = ".1"
Output: true

Constraints:

• 1 <= s.length <= 20
• s consists of only English letters (both uppercase and lowercase), digits (0-9), plus '+', minus '-', or dot '.'.

Solution: Rule checking

Time complexity: O(n)
Space complexity: O(1)

// Author: Huahua
class Solution {
public:
  bool isNumber(string s) {
    for (int i = 0; i < s.length(); ++i)
      if (s[i] != ' ') {
        s = s.substr(i);
        break;
      }
    while (!s.empty() && s.back() == ' ') s.pop_back();
    
    bool dot = false;
    bool e = false;
    bool digit = false;
    for (int i = 0; i < s.length(); ++i) {
      s[i] = tolower(s[i]);
      if (isdigit(s[i])) {
        digit = true;
      } else if (s[i] == '.') {
        if (e || dot) return false;
        dot = true;
      } else if (s[i] == 'e') {
        if (e || !digit) return false;
        e = true;
        digit = false;
      } else if (s[i] == '-' || s[i] == '+') {
        if (i != 0 && s[i - 1] != 'e')
          return false;
      } else {
        return false;
      }
    }
    return digit;
  }
};

The post 花花酱 LeetCode 65. Valid Number appeared first on Huahua's Tech Road.

Problem:

A self-dividing number is a number that is divisible by every digit it contains.

For example, 128 is a self-dividing number because 128 % 1 == 0, 128 % 2 == 0, and 128 % 8 == 0.

Also, a self-dividing number is not allowed to contain the digit zero.

Given a lower and upper number bound, output a list of every possible self dividing number, including the bounds if possible.

Example 1:

Input: 
left = 1, right = 22
Output: [1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 15, 22]

Note:

• The boundaries of each input argument are 1 <= left <= right <= 10000.

Idea:

Brute Force

Time Complexity: O(n)

Space Complexity: O(1)

Solution:

C++

// Author: Huahua
// Runtime: 3 ms
class Solution {
public:
    vector<int> selfDividingNumbers(int left, int right) {
        vector<int> ans;
        for (int n = left; n <= right; ++n) {
            int t = n;
            bool valid = true;
            while (t && valid) {
                const int r = t % 10;
                if (r == 0 || n % r)
                    valid = false;
                t /= 10;
            }
            if (valid) ans.push_back(n);
        }
        return ans;
    }
};

String

// Author: Huahua
// Runtime: 6 ms
class Solution {
public:
    vector<int> selfDividingNumbers(int left, int right) {
        vector<int> ans;
        for (int n = left; n <= right; ++n) {
            const string t = std::to_string(n);
            bool valid = true;
            for (const char c : t) {
                if (c == '0' || n % (c - '0')) {
                    valid = false;
                    break;
                }
            }
            if (valid) ans.push_back(n);
        }
        return ans;
    }
};

Related Problems:

• [解题报告] LeetCode 611. Valid Triangle Number

The post 花花酱 LeetCode 728. Self Dividing Numbers appeared first on Huahua's Tech Road.